MATH 215

Picard’s Theorem

If f(x,y) is continuous on R and if f_y exists and is continuous on R, then y(x) exists and is unique

Look for graph containing initial solution

x₀ E (-inf, x₁)…

Separable Equations

dy/dt = f(t)g(y)

• divide g(y) and integrate

• Watch for implicit (constant C)

Integrating Factor

y’ + P(x)y = g(x)

• Multiply by integrating factor mu

• Solve mu

• Product rule and integrate

Stable vs. Unstable

• Find critical points: =0

• Find the derivative + or - slope

• > * < : stable

• < * > : unstable

• > * > : semistable

Exact Equations

F_x(x,y) + F_y(x,y) dy/dx = 0

• Check F_xy = F_yx

• Integrate N w.r.t. dy

• Derivate N w.r.t. x

• solve for g_x

• Integrate g_x

• If check fails, solve integrating factor

Constant Coefficient Homogenous Linear Equations

Ay’’ + By’ + Cy = 0

• Solve with characteristic equation

• if r > 0: real, distinct roots

  • y(x) = C₁e^r₁x + C₂e^r₂x

• If r < 0: r_1,2 = a +- ib

  • y(x) = C₁e^axcos(bx) + C₂e^axsin(bx)

• If r₁=r₂: repeated

  • y(x) = C₁e^rx + C₂xe^rx

Method of undetermined coefficient

Ay’’ + By’ + Cy = f(x)

• Only works for polynomials, sin/cos, e^x

• Split into two: y_c and y_p

  • y(x) = y_c + y_p

• For guessing y_p:

  • If e^x: Axⁿe^x

  • If sin/cos: xⁿ(Acos(x) + Bsin(x))

  • If polynomial: xⁿx^s + xⁿx^s-1 + …

Spring Questions

Ay’’ + By’ + Cy = 0

my’’ + cy’ + ky = 0

• m: mass

• c: damping

• k: stiffness

• omega₀ = sqrt(k/m)
  
  

• If c = 0:

  • Never decays, roots will be pure imaginary

• If c ≠ 0:

  • If r > 0: Overdamping

    • Graph spreads outwards

  • If r < 0: Underdamping

    • Graph decays inwards

  • If r₁ = r₂: Critical Damping

    • Graph rises and falls to 0

• Complimentary solution will be used

Partial Fraction Decomposition/Integration by parts

P(x)/(x-a)(x-b) = A/(x-a) + B/(x-b)

• multiply each side, solve for A and B
  
  

\int u dv = uv - \int v du

• Find u and dv, solve for du and v

Random Derivatives

• dcos(x)/dx = -sinx

• dtan(x)/dx = sec²(x)

• dcsc(x)/dx = -csc(x)cot(x)

• dsec(x)/dx = sec(x)tan(x)

• dcot(x)/dx = -csc²(x)

• dsin^-1(x)/dx = 1/sqrt(1-x²)

• dcos^-1(x)/dx = -1/sqrt(1-x²)

• dtan^-1(x)/dx = 1/(1+x²)

• da^x/dx = a^xln(a)

6.1 Definition of the Laplace Transform

L{f(t)} = F(s) = \int^int₀ e^-st f(t) dt

• Converts time-domain → frequency-domain

• Only depends on t >= 0

• Look at formula sheet for all the possible Laplace transforms

Unit Step Function

u(t - a) = {0: t < a, 1: t >= a

• Laplace of Unit Step:

  • L{u(t - a)} = e^-as/s

Second Shifting Property (Heaviside Shift)

Laplace of a shifted function

L{f(t - a) u(t - a)} = e^-as F(s)

• Shifts graph right by a

• Turns function on at t = a

First Shifting Property

Laplace of e^-at f(t)

L{e^-at f(t)} = F(s + a)

Exponential Order

A function satisfies

• |f(t)| <= M e^ct for large t

• This ensures Laplace Transform exists for s > c

Existence Theorem

When does the Laplace Transform exist?

If f(t) is continuous and of exponential order c, then:

• L{f(t)} exists for s > c

Uniqueness Theorem

When is the inverse transform unique?

If F(s) = G(s) for all s > C, then:

• f(t) = g(t) for all t >= 0

Transforms of Derivatives

Laplace of Derivatives

If G(s) = L{g(t)}, then:

• L{g’} = sG - g(0)

• L{g’’} = s²G - sg(0) - g’(0)

• L{g’’’} = s³G - s²g(0) - sg’(0) - g’’(0)

Transform of an Integral

Laplace of integral

L{\int₀^t f(T) dT} = F(s)/s

Inverse Laplace Transform

L^-1{F(s)} = f(t)

• When using shifting property (F(s + a))

• Write quadratics as (s+a)² + b, then shift

Rules of Linearity and Product Rule

Linearity of Laplace Transforms

• L{Af + Bg} = AL{f} + BL{g}

Does L{f g} = L{f} L{g}?

• No!

Solving ODEs with Laplace (Recipe)

1. Apply Laplace to ODE (both sides)

2. Substitute initial conditions into equation

3. Solve algebraically for X(s)

4. Apply inverse Laplace to obtain x(t)

Transform of t u(t - a)

Laplace of delayed ramp

L{t u(t - a)} = e^-as(1/s² + 1/s)

Transform of Periodic Functions

Laplace transform of periodic f(t) with period P

F(s) = 1/(1-e^-Ps) \int^P₀ e^-st f(t) dt

Transfer Function

For Lx = f(t) with zero initial conditions:

• Laplace transform: A(s)X(s) = F(s)

• Transfer Function: H(s) = X(s)/F(s) = 1/A(s)

• Once H(s) is known: X(s) = H(s)F(s)

Convolution

(f * g)(t) = \int f(T) g(t- T) dT

• Produces a new function of t

Properties:

• Commutative: f * g = g * f

• Associative: (f * g) * h = f * (g * h)

• Distributive: (f + g) * h = f * h + g * h

• Scalar: c(f * g) = (cf) * g

Convolution Theorem

Laplace transform of a convolution

L{f * g} = F(s)G(s)

Convolution in ODEs

How convolution solves forced ODEs

For:

• x’’ + omega²x = f(t), x(0) = 0, x’(0) = 0

• Laplace gives:

• X(s) = 1/(s²+omega²) F(s)

Volterra Integral Equation

x(t) = f(t) + \int^t₀ g(t - T) x(T) dT

Laplace Transform:

• X(s) = F(s) + G(s)X(s)

Rectangular Pulse

P(t) = {M: a <= t < b, 0: otherwise

Using Heaviside:

• P(t) = M(u(t - a) - u(t - b))

Laplace Transform:

• L{P(t)} = M (e^-as - e^-bs)/s

Dirac Delta

Generalized function satisfying:

• \int^inf_-inf delta(t) f(t) dt = f(0)

• Zero everywhere except at t = 0

• Infinite spike with total area 1

Shifted Delta

• \int delta(t - a) f(t) dt = f(a)

• Samples the function at t = a

Laplace Transform:

• L{delta(t - a)} = e^-as

• L{delta(t)} = 1

System of ODEs

A system consists of multiple dependent variables x₁, x₂, …, x_n satisfying multiple differential equations

Example:

• x'₁ = F₁(x₁, …, x_n, t), …, x’_n = F_n(x₁, …, x_n, t)

A solution is a vector function x(t) satisfying all equations

How to convert an n-th order ODE to a system

Let:

• u₁ = y, u₂ = y’, u₃ = y’’, … , u_n = y^{(n - 1)}

Then:

• u₁’ = u₂, … , u’_n-1 = u_n, u’_n = F(u₁, …, u_n, t)

Autonomous Systems

A system where the independent variable t does not appear explicitly

Example:

• x’ = 2y - x, y’ = x

Picard’s Theorem for Systems

Existence and uniqueness

If each F_j and each partial derivative dF_j/dx_k is continuous near the initial point, then the system

• x’_j = F_j(x₁, …, x_n, t) has a unique solution near that point

Matrix Warnings

• AB ≠ BA

• AB = AC does not imply B = C

• AB = 0 does not imply A = 0 or B = 0

Matrix inverse and Determinant

AA^-1 = I = A^-1A

Only exists if A is invertible

det[[a,b],[c,d]] = ad - bc

• A is invertible if det(A) ≠ 0

• A^-1 = 1/(ad-bc) [[d, -b],[-c, a]]

Cofactor Expansion

Determinant of larger matrices

Expand along any row:

• det(A) = summationⁿ (-1)^{i + j} a_ij det(A_ij)

det(AB) = det(A)det(B)

Differentiation rules for Matrices

• (A + B)’ = A’ + B’

• (AB)’ = A’B + AB’

• (cA)’ = cA’

• (AC)’= A’C

Linear First-Order System

General Form:

• x’ = P(t)x + f(t)

• where P(t) is an nxn matrix, x and f are vectors

Homogeneous: x’ = Px

Nonhomogeneous: x’ = Px + f

Superposition Principle/Linear Independence

If x₁, …, x_n are solutions of x’ = Px, then:

• x = c₁×1 + … + c_nx_n is also a solution

• If the system is nxn and the solutions are linearly independent → general solution

Linearly independent if:

• c₁×1 + …. + c_nx_n = 0

• for all t, implies c₁ = … = c_n = 0

Fundamental Matrix

Matrix whose columns are linearly independent solutions of x’ = Px:

• X(t) = [x₁(t) x₂(t) …. x_n(t)]

• Not unique: any X(t)C with constant invertible C is also fundamental

General Solution using Fundamental matrix

x(t) = X(t)c + x_p(t)

where x_p is a particular solution of the nonhomogeneous system

X(t)c solves the homogeneous system (complimentary solution)

Solving Initial Value problems

How to apply initial condition x(t₀) = b?

Solve:

• X(t₀)c = b - x_p(t₀) for c

Then:

• x(t) = X(t)c + x_p(t)

Eigenvalue Method

How do we solve x’ = Px?

Try:

• x = ve^{lambda t}

Plug in → need:

• Pv = lambda v

So eigenvalues give exponentials, eigenvectors give directions

Eigenvalue equation

How to find eigenvalues?

Solve:

• det(P - lambda I) = 0

• degree n polynomial → n eigenvalues

How to find eigenvectors?

Solve:

• (P - lambda I) v = 0

• always has nontrivial solutions when lambda is an eigenvalue

General Solution with distinct real eigenvalues

If eigenvalues lambda₁, …, lambda_n are distinct and real:

• Corresponding eigenvectors are linearly independent

• General solution:

  • x(t) = c₁v₁e^lambda₁t + … + c_nv_ne^lambda_nt

• Fundamental Matrix:

  • X(t) = [v₁exp(lambda₁t) v₂exp(lambda₂t) … v_nexp(lambda_nt)]

  • x(t) = X(t)c

Complex Eigenvalues

If P is real and lambda = a + ib is an eigenvalue, then:

• lambda = a - ib is also an eigenvalue

How to get real solutions from complex eigenpairs?

Let:

• x₁(t) = v exp((a + ib) t)

Then:

• x₂(t) = v exp((a - ib) t) is also a solution

Real solutions:

• Re[x₁(t)], Imag[x₁(t)]

  • These are linearly independent and real-valued

Form of real solutions:

If v = p + iq, and lambda = a + ib then:

• x(t) = e^at(p cos(bt) - q sin(bt))

• y(t) = e^at(p sin(bt) + q cos(bt))

  • For each pair a+- ib, get two real solutions

  • Combine with real eigenvalue solutions to form full general solution

Cases for eigenvalues

Case 1: both lambda real and positive

• Trajectories diverge away from origin

• Behaves as an unstable node

Case 2: both lambda real and negative

• Trajectories converge to origin

• Behaves as a sink (stable node)

Case 3: real lambda with opposite signs

• Trajectories move away and toward origin

• Behaves as a saddle point (semi stable)

Case 4: Purely imaginary lambda

• lambda = +-ib

• Trajectories neither grow now decay

• Behaves as circles → center

Case 5: Complex lambda with positive real part

• a +- ib, a > 0

• Trajectories spiral outward → spiral source

Case 6: Complex lambda with negative real part

• a < 0

• Trajectories spiral inward → spiral sink

General 2D Linear System

[x’ y’] = P[x y], P = [[a,b],[c,d]]

• Autonomous (t does not appear explicitly)

• Vector field in the plane

• Assume:

  • P is 2×2

  • 2 distinct eigenvalues

  • P is nonsingular (no 0 eigenvalues)

  • Eigenvectors exist and are linearly independent

Reducing 2nd-Order to 1st-Order Form

Since M is diagonal and invertible:

• x’’ = M^-1K x = Ax

• where:

  • A = M^-1 K

Eigenvalue Method for 2nd-Order Systems

Try solution of form:

• x = v e^at

• Plug into x’’ = Ax:

  • a²ve^at = Ave^at

• Cancel e^at:

  • a²v = Av

• Thus:

  • a² is an eigenvalue of A

  • v is the corresponding eigenvector

• If lambda = -omega²:

  • a = +-i omega

  • real solutions:

    • v cos(omega t), v sin(omega t)

Zero Eigenvalue case

a² = 0 → a = 0

Solutions:

• x = v(a + bt)

• Corresponds to constant velocity motion

General Solution (Distinct Negative lambda)

If lambda are:

• -omega₁² > -omega₂² > … > -omega_n²

Then:

• x(t) = sumⁿ v_j (a_j cos(omega_j t) + b_j sin(omega_j t))

• If one lambda = 0:

  • x(t) = v₁(a₁+b₁t) + sum_j=2 v_j (a_j cos(omega_j t) + b_j sin(omega_j t))

Forced Oscillations

System:

• x’’ = Ax + F cos(omega t)

Try x_p:

• x_p= c cos(omega t)

• Plug in:

  • -omega²c = Ac + F

• Solve:

  • c = -(A + omega² I)^-1 F

  • valid only if -omega² is not an eigenvalue of A

  • If it is, resonance → try:

    • x_p= t (C sin(omega t) + D cos(omega t))

Algebraic vs. Geometric Multiplicity

• Algebraic multiplicity (AM):

  • Number of times the eigenvalue appears as a root of the characteristic polynomial

• Geometric multiplicity (GM):

  • Number of linearly independent eigenvectors associated with that eigenvalue

• Always:

  • 1 <= GM <= AM

  • If GM = AM → eigenvalue is complete

  • If GM < AM → eigenvalue is defective

Complete Eigenvalues

GM = AM, you have enough eigenvectors to form full general solution:

• x(t) = sumⁿ c_jv_je^lambda_jt

As long as you have n independent eigenvectors, the system behaves like the distinct eigenvalue case

Generalized Eigenvectors

A vector v_k satisfying:

• (A - lambda I)^k v_k = 0, (A - lambda I)^k-1 v_k ≠ 0

For a defective eigenvalue of multiplicity 2:

• FInd eigenvector v₁ such that:

  • (A - lambda I)v₁ = 0

• Find generalized eigenvector v₂ such that:

  • (A - lambda I)v₂ = v₁

General Solution for Defect 1 (AM = 2, GM = 1)

Solutions are:

• x₁(t) = v₁e^{lambda t}

• x₂(t) = (v₂ + tv₁) e^{lambda t}

General Solution

• x(t) = c₁v₁e^{lambda t} + c₂(v₂ + tv₁) e^{lambda t}

Example: AM = 3, GM = 1

Solutions:

• x₁ = v₁e^{lam t}

• x₂ = (v₂ + tv₁) e^{lam t}

• x₃ = (v₃ + tv₂ + t²/2 v₁) e^{lam t}

Matrix Exponential

Use the Taylor series, replacing powers of a number with powers of a matrix:

• e^Pt = I + Pt + (Pt)²/2! + (Pt)³/3! + …

General Solution using Matrix Exponential:

• x(t) = e^Pt b

• where b = x(0)

• Thus e^Pt is a fundamental matrix solution

Does e^A+B = e^Ae^B?

• Only if AB = BA

• Else:

  • e^A+B ≠ e^Ae^B

Diagonal Matrices

If D = [[a, 0],[0, b]]

Then e^D = [[e^a, 0],[0, e^b]]

• General: exponentiate each diagonal entry

Nilpotent Matrices

What if a matrix B^k = 0 for some k?

Then the exponential truncates:

• If B² = 0:

  • e^B = I + B

• If B³ = 0:

  • e^B = I + B + B²/2

Similarity Transform Property

If A is diagonalizable:

• A = EDE^-1

Use identity:

• e^{BAB^-1} = Be^AB^-1

• e^tA = Ee^tDE^-1

where:

• e^tD = diag(e^lam₁t, … , e^lam_nt)

Using a Known Fundamental Matrix to get e^tA

If X(t) is any fundamental matrix solution:

• e^tA = X(t) [X(0)]^-1

• Because e^0A = I

General Form of a nonhomogeneous 1st-Order System

x’(t) = Ax(t) + f(t)

where:

• A = constant matrix

• f(t) = forcing vector

• Normally solve with methods like integrating factor

When integrating factor fails

d/dt (e^{\int P(t) dt} ≠ P(t) e^{\int P(t) dt}

• Matrix multiplication is not commutative, so scalar trick breaks

• Integrating factor only works if P is a constant, not P(t)

Eigenvector Decomposition Method

If A has n independent eigenvectors v₁, …, v_n:

• x(t) = sum v_k E_k(t)

Decompose forcing:

• f(t) = sum v_k g_k(t)

Then each scalar equation becomes:

• E’_k = lam_k E_k + g_k

Solve with integrating factor:

• E_k = e^lam_kt (\int e^{-lam_k s} g_k ds + a_k)

Works when A has n linearly independent eigenvectors (diagonalizable)

If defective → must use generalized eigenvectors

Undetermined Coefficients for Systems

Guess:

• x_p= ae^at + bte^at + …

• If guess conflicts with complementary solution → multiply by t

Variation of Parameters (General A(t))

Given:

• x’ = Ax + f

Try:

• x_p= Xu

• Solve for u’ and integrate

Thus:

• x_p= X(t) \int X^-1 f ds

2nd-Order Systems

• Undetermined Coefficients

  • Guess vector form

• Eigenvector Decomposition

  • write:

    • x(t) = sum v_kE_k

  • Then each scalar equation becomes:

    • E’’_k = lam_kE_k + g_k
      
      

Autonomous Systems and Phase Plane

Autonomous System:

• x’ = f(x,y)

• y’ = g(x,y)

Phase Plane / Phase Portrait

• Plot of (x(t), y(t))

• Vector field arrows show the direction of motion

• Trajectories follow the arrows

Critical Point (Equilibrium)

• A point (x₀, y₀) where

  • f(x₀, y₀) = 0

  • g(x₀, y₀) = 0

• Corresponds to constant solution x = x₀, y = y₀

Linearization

Approximate nonlinear system near a critical point using a linear system

Change of variables:

• u = x - x₀

• v = y - y₀

Jacobian Matrix:

• J(x₀, y₀) = [[f_x, f_y], [g_x, g_y]]

Linearized System:

• [u’ v’] = J(x₀, y₀) [u v]

Interpretation:

• Near the critical point, nonlinear behavior = linear behavior

• Works only when Jacobian is invertible

Isolated Critical Points and Almost Linear Systems

Isolated Critical Point:

• Only one critical point in some neighborhood

Almost Linear System:

• Critical point isolated

• Jacobian Invertible

• therefore, nonlinear system behaves like its linearization nearby

Classification via Eigenvalues

Use eigenvalues of the Jacobian at the critical point

Real, both positive:

• Source / Unstable Node

• Trajectories move away

Real, both negative:

• Sink / Stable Node

• Trajectories move toward point

Real, opposite signs:

• Saddle

• Always unstable

Complex eigenvalues, Re > 0:

• Spiral Source

• Outward spiraling

Complex eigenvalues, Re < 0:

• Spiral Sink

• Inward spiraling

Purely imaginary eigenvalues:

• Center (linear case)

• Closed orbits, no spiraling

• Nonlinear case: linearization inconclusive

Issue with Centers

• Linearization gives purely imaginary eigenvalues → center

• But nonlinear terms may push trajectories inward or outward

• Must analyze system more deeply

Conservative Systems

Form:

• x’’ + f(x) = 0

Convert to 1st-order:

• x’ = y

• y’ = -f(x)

Energy Integral (Hamiltonian)

• ½ y² + \int f(x) dx = C = H(x,y)

Implications

• Energy conserved → trajectories lie on level curves

• No spirals (cannot approach or diverge)

• Only 2 types of critical points:

  • Saddle if f’(x₀) < 0

  • Center if f’(x₀) > 0

Jacobian:

• J = [[0, 1], [-f ‘(x), 0]]

Eigenvalues:

• lam = +- sqrt(-f’(x₀))

Pendulum Equation

Equation:

• theta’’ + g/L sin(theta) = 0

1st-order system:

• theta’ = omega

• omega’ = -g/L sin(theta)

Critical points:

• (theta, omega) = (k pi, 0)

Classification:

• Even multiples of pi: centers

  • pendulum hanging straight down

• Odd multiples of pi: saddles

  • pendulum inverted at the top

Phase Portrait Features:

• Closed loops around centers (oscillations)

• Separatrices connecting saddles

• Large energy → full rotations

Stability Definitions

Stable:

• Trajectories stay near the critical point

Asymptotically Stable:

• Trajectories approach the point as t → infinity

Unstable:

• Some trajectories move away

Partial Fraction Decomp

Case 1: Distinct Linear Factors

• Q(s) = (s-a)(s-b)(s-c)

• Then:

  • P(s)/Q(s) = A/s-a + B/s-b + C/s-c

Case 2: Repeated Linear Factors

• Q(s) = (s-a)ⁿ

• Then:

  • P(s)/Q(s) = A₁/s-a + A₂/(s-a)² + … + A_n/(s-a)ⁿ

Case 3: Irreducible Quadratic Factors

• Q(s) = s² + bs + c

• If it does not factor, then numerator must be linear

  • As + B/(s² + bs + c)